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(II) A centrifuge rotor rotating at 9200 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m\(\cdot\)N. If the mass of the rotor is 3.10 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

Short Answer

Expert verified
The rotor turns 480.57 revolutions and takes 6.27 seconds to come to rest.

Step by step solution

01

Find the Moment of Inertia

The given rotor can be approximated as a solid cylinder. The moment of inertia \( I \) for a solid cylinder is given by \( I = \frac{1}{2} m r^2 \). Substituting the given values, \( m = 3.10 \, \text{kg} \) and \( r = 0.0710 \, \text{m} \), we calculate:\[I = \frac{1}{2} \times 3.10 \, \text{kg} \times (0.0710 \, \text{m})^2 = 0.00782 \, \text{kg} \cdot \text{m}^2.\]
02

Find the Initial Angular Velocity

The initial angular velocity \( \omega_0 \) can be found by converting the rotational speed from rpm to rad/s. Since \( 9200 \, \text{rpm} = 9200 \times \frac{2\pi}{60} \, \text{rad/s} \), we have:\[\omega_0 = 9200 \times \frac{2\pi}{60} = 962.11 \, \text{rad/s}.\]
03

Calculate Angular Acceleration

Use the relation between torque \( \tau \) and angular acceleration \( \alpha \):\[ \tau = I \alpha. \]Substitute the values \( \tau = 1.20 \, \text{m} \cdot \text{N} \) and \( I = 0.00782 \, \text{kg} \cdot \text{m}^2 \) to find \( \alpha \):\[1.20 = 0.00782 \alpha \Rightarrow \alpha = \frac{1.20}{0.00782} \approx 153.46 \, \text{rad/s}^2.\]
04

Find the Time to Come to Rest

The rotor comes to rest when its final angular velocity \( \omega_f = 0 \). Use the equation:\[ \omega_f = \omega_0 + \alpha t \]Solving for \( t \):\[0 = 962.11 + (-153.46) t \Rightarrow t = \frac{962.11}{153.46} \approx 6.27 \, \text{s}.\]
05

Determine Total Revolutions

Use the equation for angular displacement \( \theta \) in terms of \( \omega_0 \), \( \alpha \), and \( t \):\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2. \]Substitute the known values \( \omega_0 = 962.11 \, \text{rad/s} \), \( \alpha = -153.46 \, \text{rad/s}^2 \), and \( t = 6.27 \, \text{s} \):\[\theta = 962.11 \times 6.27 + \frac{1}{2} \times (-153.46) \times (6.27)^2 \approx 3018.48 \, \text{rad}. \]Convert radians to revolutions:\[ \frac{3018.48}{2\pi} \approx 480.57 \text{ revolutions}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around an axis. It tells us the rate of rotation in terms of angle turned per unit time. In this exercise, the initial angular velocity of the centrifuge rotor is given in revolutions per minute (rpm), which is a common way to express rotational speed in mechanical systems.

To convert from rpm to radians per second (rad/s), which is the SI unit, we use the conversion factor of \(2\pi/60\) since one full revolution is \(2\pi\) radians and there are 60 seconds in a minute. So, \(9200 \, \text{rpm} = 9200 \times \frac{2\pi}{60} \, \text{rad/s} \). This conversion tells us how fast the rotor initially spins in terms understood by physics equations involving rotational motion.
  • Helps predict how a rotating object will behave after a certain time.
  • Important in calculating rotational kinetic energies and other parameters.
Torque
Torque is the rotational analog of force. It measures the tendency of a force to rotate an object about an axis. In this problem, the frictional torque plays a crucial role in bringing the spinning rotor to a stop.

The formula \( \tau = I \alpha \), where \(\tau\) is torque, \(I\) is the moment of inertia, and \(\alpha\) is angular acceleration, underscores its importance in rotational motion. Here, the frictional torque of \(1.20 \, \text{m} \cdot \text{N}\) acts oppositely to the rotor's spin, causing it to decelerate and eventually halt. This frictional force provides the necessary change in motion despite the initial high-speed rotation.
  • Greater torque can produce a higher rate of rotational acceleration.
  • Key in applications like car engines and balancing wheels.
Moment of Inertia
The moment of inertia is a property of a rotating body that represents its resistance to change in rotational motion. It depends on the mass distribution of the object relative to the axis of rotation.

For a solid cylinder, often used in approximations like this exercise, the moment of inertia formula is given by \( I = \frac{1}{2} m r^2 \). Given the rotor's mass and radius, the calculation yields \( 0.00782 \, \text{kg} \cdot \text{m}^2\).
  • Higher moment of inertia means more effort is required to change the object's rotational speed.
  • Critical in designs of rotational devices to manage energy use efficiently.
Angular Acceleration
Angular acceleration is defined as the rate of change of angular velocity with respect to time. It tells us how quickly the object is speeding up or slowing down in its rotation.

The relation \( \tau = I \alpha \) allows us to compute the angular acceleration given the known torque and moment of inertia in this exercise. From there, we can determine how long it will take for the rotor to come to a stop, and through how many rotations it will travel before stopping.
  • A key concept in understanding rotational motion and designing mechanical systems.
  • Affects everything from spinning tops to complex machinery such as turbines.

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Most popular questions from this chapter

(I) A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?

Water drives a waterwheel (or turbine) of radius \(R = 3.0 m\) as shown in Fig. 8-66. The water enters at a speed \(\upsilon_1 = 7.0 m/s\) and exits from the waterwheel at a speed \(\upsilon_2 = 3.8 m/s\). \((a)\) If 85 kg of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? \((b)\) What is the torque the water applies to the waterwheel? \((c)\) If the water causes the waterwheel to make one revolution every 5.5 s, how much power is delivered to the wheel?

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(II) \((a)\) What is the angular momentum of a figure skater spinning at 3.0 rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 15 cm, and a mass of 48 kg? \((b)\) How much torque is required to slow her to a stop in 4.0 s, assuming she does \(not\) move her arms?

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